Chapter 8.6, Problem 56E

To determine

To find: the probability of the event.

Answer to Problem 56E

The probability of selected all three numbers such that two numbers are odd and the number is 2, 4, or 6 is $\frac{3}{40}$ .

Explanation of Solution

Given information:

A random number generator selects three numbers from 1 through 10.

One number is 2, 4, or 6.

Other two numbers are odd.

Calculation:

Consider the 1 to 10 numbers. Three numbers randomly selected from 1 to 10 numbers.

There are total 3 possibilities which may be 2, 4, or 6, and the other two numbers are odd, so probability of selecting three such numbers is,

  $\begin{array}{c}P\left(E\right)=\frac{3}{10}\cdot \frac{5}{10}\cdot \frac{5}{10}\\ =\frac{75}{1000}\\ =\frac{3}{40}\end{array}$

Hence, the probability of selected all three numbers such that two numbers are odd and the number is 2, 4, or 6 is $\frac{3}{40}$ .